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Using Kepler's third law, determine the period

  1. Mar 25, 2015 #1
    1. The problem statement, all variables and given/known data
    If there were a planet three times farther from the sun than the Earth is, how long would it have taken this hypothetical planet to orbit the Sun? Assume the orbit is a circle.

    2. Relevant equations
    Kepler's 3rd Law ##= (\frac{r_1}{r_2})^3 = (\frac{T_1}{T_2})^2 ##

    3. The attempt at a solution
    ##= (\frac{3r_E}{r_E})^3 = (\frac{T}{1})^2 ##
    T = 3 years Correct answer: T = 5.2 years
    Where is my blunder? Thank you all so much.
     
  2. jcsd
  3. Mar 25, 2015 #2

    gneill

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    Staff: Mentor

    Show the math one step at a time.
     
  4. Mar 25, 2015 #3
    I just found out my error. ##3^3 = 27##. I accidentally wrote 9 instead of 27.
    Thanks for helping Gniell
     
  5. Mar 25, 2015 #4
    I have one more question...
    Part B of this question asks if I can use the data to solve for the mass of the hypothetical planet. The answer is no. "No mass data can be calculated from this relationship, because the relationship is mass- independent. Any object at the orbit radius of 3 times the Earth’s orbit radius would have a period of 5.2 years, regardless of its mass. "

    The "...regardless of its mass" statement seems counter-intuitive to me. What if that planet turned out to be a super-giant red star much greater in mass than the Sun? Then the Sun would orbit the red star right? How can the red star still have an orbital period of 5.2 years? Thank you.
     
  6. Mar 25, 2015 #5

    gneill

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    Staff: Mentor

    The "...regardless of its mass" statement needs to be taken in the context of the basic assumption of Kepler's Laws where the masses of all the planets are considered to be much less than that of the primary (the Sun).

    So I guess one might say that the mass of the object doesn't matter as long as it is insignificant :smile:
     
  7. Mar 25, 2015 #6
    Thank you
     
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